A triangle has corners at (2 , 2 ), ( 5, 6 ), and ( 1, 4 )#. What are the endpoints and lengths of the triangle's perpendicular bisectors?
1 Answer
Each of the 3 segments have one endpoint at the midpoint of the given points and the other endpoint at
Explanation:
Repeating the points
Midpoints
Slopes of segments (
Now to simplify the work that remains to do, be noted that the line perpendicular to a side meets one of the other sides (or both at the same time when it intercepts a vertex of the triangle). When it meets just one of the two other sides, which side is this? Answer: the LONG side (although that is intuitive it can be proved). Then we need to know the lengths of the sides of the triangle.
=>
So
line 1 perpendicular to AB meets side BC
line 2 perpendicular to BC meets side AB
line 3 perpendicular to AC meets side AB
We need the equations of the lines in which the sides AB and BC lay and the equations of the 3 perpendicular lines
Equation of the line that supports side:
Equation of the line (passing through midpoint) perpendicular to side:
Finding the interceptions on sides AB and BC
Combining equations [b] and [1]
{y=(x+7)/2
{y=(-3x+26.5)/4 =>(x+7)/2=(-3x+26.5)/4 =>2x+14=-3x+26.5 =>5x=12.5 =>x=2.5
-> y=(2.5+7)/2 =>y=4.75 We've found
R(2.5,4.75)
The distance betweenM_(AB) and R is
d1=sqrt((2.5-3.5)^2+(4.75-4)^2)=sqrt(1+.5625)=1.25
Combining equations [a] and [2]
{y=(4x-2)/3
{y=-2x+11 =>(4x-2)/3=-2x+11 =>4x-2=-6x+33 =>10x=35 =>x=3.5
-> y=-2*3.5+11 =>y=4 We've found
S(3.5,4)
We can find the distance betweenM_(BC) andS :
d2=sqrt((3.5-3)^2+(4-5)^2)=sqrt(1.25)=5*sqrt(5)
Combining the equations [a] and [3]
{y=(4x-2)/3
{y=(-x+7.5)/2 =>(4x-2)/3=(-x+7.5)/2 =>8x-4=-3x+22.5 =>11x=26.5 =>x=53/22
-> y=(-53/22+15/2)/2=(-53+165)/44 =>y=28/11 We've found
T(53/22, 28/11)
The distance betweenM_(CA) and T is
d3=sqrt((53/22-3/2)^2+(28/11-3)^2)=sqrt((10/11)^2+(5/11)^2)=sqrt(125)/11=5*sqrt(5)/11